Multiply the integrand's numerator and denominator by [tex]1-\sin x[/tex], then simplify:
[tex]\displaystyle\int\frac{\mathrm dx}{1+\sin x}=\int\frac{1-\sin x}{1-\sin^2x}\,\mathrm dx=\int\frac{1-\sin x}{\cos^2x}\,\mathrm dx=\int(\sec^2x-\sec x\tan x)\,\mathrm dx[/tex]
You should be familiar with
[tex]\dfrac{\mathrm d}{\mathrm dx}\tan x=\sec^2x[/tex]
[tex]\dfrac{\mathrm d}{\mathrm dx}\sec x=\sec x\tan x[/tex]
so that
[tex]\displaystyle\int\frac{\mathrm dx}{1+\sin x}=\tan x-\sec x+C[/tex]
